Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int NMAX = 300000;
const double eps = 1e-6;
double sy, a, b;
int n, q;
double L[NMAX], R[NMAX];
double sum[NMAX];
int dcmp(double x, double y) {
  if (fabs(x - y) < eps) return 0;
  if (x > y) return 1;
  return -1;
}
int bin_search(double x) {
  int l = 1, r = n, ret = 0;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (dcmp(x, R[mid]) >= 0) {
      ret = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  return ret;
}
int main() {
  scanf("%lf%lf%lf", &sy, &a, &b);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf%lf", &L[i], &R[i]);
    sum[i] = sum[i - 1] + R[i] - L[i];
  }
  scanf("%d", &q);
  while (q--) {
    double x0, y0, x1, x2;
    scanf("%lf%lf", &x0, &y0);
    if (dcmp(x0, a) == 0)
      x1 = a;
    else {
      double k = (y0 - sy) / (x0 - a);
      x1 = a - sy / k;
    }
    if (dcmp(x0, b) == 0)
      x2 = b;
    else {
      double k = (y0 - sy) / (x0 - b);
      x2 = b - sy / k;
    }
    int r1 = bin_search(x1);
    int r2 = bin_search(x2);
    double ans = (sum[r2] - sum[r1]);
    if (r1 + 1 <= n && dcmp(L[r1 + 1], x1) < 0) ans -= x1 - L[r1 + 1];
    if (r2 + 1 <= n && dcmp(L[r2 + 1], x2) < 0) ans += x2 - L[r2 + 1];
    printf("%.15lf\n", (b - a) * ans / (x2 - x1));
  }
  return 0;
}